\hspace{-16}$If $\bf{a_{1}+a_{2}.2!+a_{3}.3!+......+a_{n}.n!=695}$. Then $\bf{a_{4}=}$\\\\ If $\bf{0\leq a_{k}\leq k}$ and $\bf{n!=n.(n-1).(n-2)........3.2.1}$
3max value of n can be 5. i got this -
a1 = 1.
a2 = 2.
a3 = 3.
a4 = 3.
a5 = 5.
these satisty 0≤ak≤k
1057this can be done by unit's digit method....
for all terms of 5! and above the unit's digit will always be 0....
so a1+2a2+6a3+24a4 shud have 5 in its unit's digit.... since the last 4 terms are even a1 has to be odd....
and then what h4hemang said....
